Ridiculous Fermi Questions

There is a new book that I am in love with. It is called "What If? Serious Scientific Answers to Absurd Hypothetical Questions" by Randall Munroe. If you have wondered if you could build a jetpack out of downward firing machine guns or what would happen if you threw a baseball at the speed of light, this book is for you.

Book Link

I decided to look at some of their pages to find a Fermi question. I think I found a good one. Ink molecules.

"Suppose you were to print, in 12 point text, the numeral 1 using a common cheap ink-jet printer. How many molecules of the ink would be used? At what numerical value would the number printed approximately equal the number of ink molecules used?"

Hyperbole and a Half. Highly recommended.

At this point, I stopped reading the page. I am now going to attempt to figure this one out on my own. Normally, when I want to find the number of particles in a substance, I use the following equation:

# of Particles = Mass of Object * Ratio of Mols/Unit Mass * Ratio of Particles/Mol

I therefore want to find out 2 things. (1) The ratio of particles/mass of printer ink and (2) the mass of printer ink used to print the number 1.

1. Ink Particles/Mass

Ok, so the first number that will be useful here is called Avogadro's Number, which is the ratio of particles in one mol of particles. We use mols instead of pounds or kilograms because different particles weigh different amounts. A hydrogen atom has a mass of 1.008 atomic units, while a carbon atom has a mass of 12.01 atomic units. Avogadro found that 1 mol of hydrogen is 1.008 kilograms and that 1 mol of carbon is 12.01 kilograms. In fact, he found that for any particle of atomic mass m, one mol of that particle has mass m kilograms. This number is 6.023*10^23 particles/mol.

Think of how much guacamole you could make with that many avocados...

Next we need to figure out the chemical composition of ink. It appears that ink can have many different chemicals in it, but mostly ink is made of carbon suspended in oils and water. Water has an atomic mass of 16 and carbon has an atomic mass of 12, so if we assume a 50/50 mix, this comes out to an atomic mass of 14. This means that 1 mol, or 6.023*10^23 particles of "ink" has a mass of 14 kilograms.

2. The Mass of Printer Ink Used

Sometimes I'll start a sentence and I don't even know where it's going. I just hope I find it along the way.

The size of fonts is typically specified by "points", which it turns out are very specific. 1 point is equal to exactly 1/6th of an inch. I am using metric units, so 1 point is 352.78*10^-6 meters, and a point refers to the vertical distance between lines on a page. So 12 point font has a vertical size of 4.2*10^-3 meters.

"The metric system is the tool of the Devil! My car gets
40 rods to the hogs-head, and that's the way I likes it!" - Abe Simpson

There isn't really a measurement for the width of letters, so I decided to look at giant number and guess the size. It looks like this 1 is twice as tall as it is wide, but this looks like a REALLY wide 1. So I am going to estimate that for normal print, the width of a number 1 is 1/5th the height of the number, which is .84*10^-3 meters.

The thickness is another difficult number to find. I found many different estimates about the thickness of ink on paper depending on the printing process. Most estimates are between 1 and 20 microns (a micron is 10^-6 meters). I am going to estimate that printed ink is 10 microns thick.

Now we know the dimensions of the number 1 in 12 point font. We can calculate the volume of ink used to print the number 1.

Volume = Length * Width * Height

Volume = (4.2*10^-3m)*(8.4*10^-4m)*(10*10^-6m)

Volume = 3.6*10^-11 meters cubed

Now to find the mass of printer ink, we need to find the density. The density of water is 1000 kg/m^3 and the density of carbon is 1800 kg/m^3. So I am going to estimate that the density of printer ink is about 1400 kg/m^3.

Mass = Volume * Density

Mass = (3.6*10^-11 m^3)*(1400 kg/m^3)

Mass = 5*10^-8 kg

Now that we have the mass of printer ink and the atomic mass of printer ink, we can find the number of particles.

# of Particles = Mass of Object * Ratio of Mols/Unit Mass * Ratio of Particles/Mol

# of Particles of Ink = (5*10^-8 kg)*(1/14 mol/kg)*(6.023*10^23 particles/mol)

# of Particles of Ink = 2*10^15 molecules.

My estimate is that there are roughly 10^15 molecules per letter typed in 12 point font.

Now onto the next part: At what numerical value would the number printed approximately equal the number of ink molecules used?

10000000000000000 (10^15) is the number of particles per letter typed in 12 point font, but in typing this number, I now have 16 digits instead of 1. This means that there are 16 times as many particles here. And now if we multiple 10000000000000000 (10^15) by 16, we get 160000000000000000. But that just added another digit, or another 10^15 particles. If we add one more multiple of 10^15, we have 170000000000000000, or 1.7*10^16.

So in total, my guess is that there are 10^15 molecules of ink that make up the digit "1" in 12 point font, and that 170000000000000000 (or 1.7*10^16) printed in 12 point font has roughly 170000000000000000 molecules of ink.

Time to check with Randall.

How I imagine Randall would react to this answer.

His first estimate without doing any calculations is that there are about 10^17 molecules per letter and the number that has the same number of particles as numerical value is between 10^18 and 10^19. So far my estimate is not too bad. We differ by a factor of 100.

When he looks at the actual numbers, he found that he overshot by a factor of around 100, which puts his estimates very close to mine. This is rather surprising since he went about finding this answer VERY differently (I'll let you see for yourself).

By the way, since we've been talking about XKCD and moles, if you've ever wondered what a mol of moles would look like, look no further!

Spoiler: It's a lot of moles